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1. I'm really confused as how to answer 1c, the answer is 30/49 can someone explain
I'm really confused as how to answer 1c, the answer is 30/49 can someone explain

This has slipped through the cracks as it was not posted in the Maths forum.
3. (Original post by Kevin De Bruyne)

This has slipped through the cracks as it was not posted in the Maths forum.
According to my teacher you do 7/7 X 5/7 x 6/7 which doesn't make sense to me
According to my teacher you do 7/7 X 5/7 x 6/7 which doesn't make sense to me
It's very similar to a red/yellow ball example - if you are pulling them out without replacement and you have 3 of each, what is the probability of getting all reds? 3/6 * 2/6 * 1/6
5. (Original post by Kevin De Bruyne)
It's very similar to a red/yellow ball example - if you are pulling them out without replacement and you have 3 of each, what is the probability of getting all reds? 3/6 * 2/6 * 1/6
Surely without replacement the denominator will also decrease by one as every time you take a ball out, the number of total balls left is reduced by one?
Surely without replacement the denominator will also decrease by one as every time you take a ball out, the number of total balls left is reduced by one?
What the **** am I posting..
7. (Original post by Kevin De Bruyne)
Correct, silly mistake by me..!

Which would mean your teacher has slipped too
The answer is also in the book tho
The answer is also in the book tho

Consider the first woman - she gives birth, and that accounts for one of the days of the week.

Now the second woman: To be on a different day to the first she only has a choice of 6 out of the 7 days. So, probabilty it's a different day is 6/7.

Now the third: Two days are taken, so she only has a "choice" of 5 out of the 7.

Hence probabilty of all different days is (6/7) x (5/7) = 30/49

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